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A particle of mass m executes simple harmonic motion with amplitude a and angular frequency omega. At a displacement x from the equilibrium position, the particle receives an impulse J in the direction of its velocity. Which of the following statements are CORRECT?

1. The amplitude of the new SHM increases.
2. The time period of the new SHM remains unchanged.
3. The new amplitude is sqrt((J/(m*omega) + sqrt(a² - x²))² + x²).
4. The frequency of the new SHM may increase.

Correct answer: The amplitude of the new SHM increases.

Solution

The angular frequency omega = sqrt(k/m) depends only on the spring constant and mass, so the time period and frequency do not change. The added impulse increases velocity and hence energy, so the new amplitude is larger.

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